【最短路】poj3268 Silver Cow Party

版权声明：转载标注来源喔~ https://blog.csdn.net/iroy33/article/details/89403919

题意：有向图，所有的牛要到X，还要回来，给出路的长度，问往返最长时间的牛的耗时

思路：和常规的不一样的是要算所有点到X的最短路+X到所有点的最短路

后者是常规操作，前者的解决措施为:反向建图跑以X为源点的最短路，这可太秀了

所有点到X的最短路：反向建图跑以X为源点的最短路

//ÑùÀý¹ýÁË£¬poj±ÀÁË

#include<cstdio>
#include<queue>
#include<vector>
#include<algorithm>
#include<cstring>
using namespace std;
typedef pair<int, int> pii;
const int N = 1007;
const int M = 1e5 + 10;
const int inf = 0x3f3f3f3f;
struct Edge
{
    int v, w, next;
    Edge() {}
    Edge(int v, int w, int next) : v(v), w(w), next(next) {}
}edge1[M],edge2[M];
int cnt1, cnt2;
int n, m, x;
int dist1[N], dist2[N];
int head1[N], head2[N];
void AddEdge(int u, int v, int w)
{
    edge1[cnt1] = Edge(v, w, head1[u]);
    head1[u] = cnt1++;
    edge2[cnt2] = Edge(u, w, head2[v]);
    head2[v] = cnt2++;
}
void dijkstra(int s)
{
    for(int i = 1; i <= n; ++i)
    {
        dist1[i] = inf;
        dist2[i] = inf;
    }

    priority_queue<pii, vector<pii>, greater<pii> > pq;
    dist1[s] = 0;
    pq.push(pii(dist1[s], s));
    while(!pq.empty())
    {
        pii p = pq.top(); pq.pop();
        int u = p.second;
        if(dist1[u] < p.first) continue;
        for(int i = head1[u]; i != -1; i = edge1[i].next)
        {
            int v = edge1[i].v;
            if(dist1[v] > dist1[u] + edge1[i].w)
            {
                dist1[v] = dist1[u] + edge1[i].w;
                pq.push(pii(dist1[v], v));
            }
        }
    }
    dist2[s] = 0;
    pq.push(pii(dist2[s], s));
    while(!pq.empty())
    {
        pii p = pq.top(); pq.pop();
        int u = p.second;
        if(dist2[u] < p.first) continue;
        for(int i = head2[u]; i != -1; i = edge2[i].next)
        {
            int v = edge2[i].v;
            if(dist2[v] > dist2[u] + edge2[i].w)
            {
                dist2[v] = dist2[u] + edge2[i].w;
                pq.push(pii(dist2[v], v));
            }
        }
    }

}
int main()
{
    scanf("%d%d%d", &n, &m, &x);
    memset(head1,-1,sizeof(head1));
    memset(head2,-1,sizeof(head2));
    for(int i = 1; i <= m; ++i)
    {
        int a, b, t;
        scanf("%d%d%d", &a, &b, &t);
        AddEdge(a, b, t);
    }
    dijkstra(x);
    int ans = 0;
    for(int i = 1; i <= n; ++i)
    {
        int tmp = dist1[i] + dist2[i];
        ans = max(tmp, ans);
    }
    printf("%d\n",ans);
    return 0;

}